B1=18
q=2/3
S=b1/(1-q)=18:(1-2/3)=18:1/3=18*3=54
16x^2-12x-4x+3-(16x^2-16x+4)=-1
Cos(a-π/6)*cos(a+π/6)=1/2(cos(a-π/6-a-π/6)+cos(a-π/6+a+π/6))=
=1/2(cos(-π/3)+cos2a)=1/2*cosπ/3+1/2cos2a=1/2*1/2+1/2*cos2a=
=1/4+1/2*cos2a
(1 + tg²a)•cos⁴a + sin²a = 1
(1 + sin²a/cos²a)•cos⁴a = 1 - sin²a
(1 + sin²a/cos²a)•cos⁴a = cos²a
1 + sin²a/cos²a = 1/cos²a
cos²a/cos²a + sin²a/cos²a= 1/cos²a
(sin²a + cos²a)/cos²a = 1/cos²a
1/cos²a = 1/cos²a
P.s., вместо ctg²a должен быть tg²a, иначе тождество неверно.